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QUESTION PAPER CODE 30/1/3
EXPECTED ANSWERS/VALUE POINTS
SECTION - A
Q.No.
1.
25
o
2.
–9
4
Marks
3. 1:3
4.
21
26
1×4 = 4 m
SECTION - B
In Δs TPC and TQC
TP  TQ
5.
TC  TC
 1   2 (TP and TQ are equally
inclined to OT)







1m
 Δ TPC  Δ TQC
 PC  QC and  3   4
 3   4  180 o   3   4  90 o 

 OT is the right bisector of PQ

½m
But
6.
7.
½m
The given A.P. is 6, 13, 20, ---, 216
 ABQ 
Let n be the number of terms, d = 7, a = 6
½m
 216 = 6 + (n – 1) .7 
½m
n = 31
 Middle term is 16th
½m
 a16 = 6 + 15 × 7 = 111
½m
1
 AOQ  29o
2
 ATQ  180o –  ABQ   BAT   180 o – 119o  61o
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8.
ABC is right triangle
2
2
2
AC = BC + AB

2
2
AB 2  5 – 2   2  2   25  AB  5 

2
2
2
BC 2  2  2   t  2   16  t  2 


2
2
2
2
AC  5  2   2 – t   49  2 – t 


49  2 – t   41  t  2 
2
t  22 – 2 – t 2 
2
8
4  2t  8  t  1
9.






1
2  3  8 K  2  3K  3
K 1
4
 K1
5
 Required ratio = 1 : 5
10.
1m
Let P divide AB in the ratio of k : 1
2K
1m
½m






1m
½m
The given quadratic equation can be written as
(9x2 – 6b2x + b4) – a4 = 0
2
½m
2
or (3x – b2) – (a2) = 0 or (3x – b2 + a2) (3x – b2 – a2) = 0

x
b 2 – a 2 b2  a 2
,
3
3
1m
½m
SECTION - C
11.
Let OA = OB = r

40 
22 r 22
 
 r  r  280  40r
7 2 7
r7




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
 1 22 7 7 1 22

shaded area   
   
 7  7  cm 2
2 7 2 2 2 7

1m
5
385
1

  77   or
cm 2  96 cm 2
4
4
4

12.
1m
Volume of solid wooden toy
5
2 22 7 7 7 1 22 7 7
 
    
  h
6
3 7 2 2 2 3 7 2 2
1001 22 7 7
or

  7  h 
6
7 2 2
166

7 h 
1001 7
 13
22  7

h  6 cm
Area of hemispherical part of toy   2  22  7  7  cm 2
7 2 2

 77 cm 2






½+½ m





½m
Cost of Paenting = Rs. (77 × 10) = Rs. 770
13.
½m
ARQ ~ ADC

½m
x 4

 x  2
6 12
½m
82  4 2  4 5
½m
QC 

Total surface area of frustum PQCB 


2
2
 π 6  2   4 5  6   2  

1m
22
32  2.236  40  22 111.552   22  15.936
7
7
 350.592
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



1m
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14.
Total surfacearea of solid cuboidal block
= 2 (15 × 10 + 10 × 5 + 15 × 5) cm2 = 550 cm2
Area of two circular bases = 2 
1m
22 7 7
   77 cm 2
7 2 2
Area of curved surface of cylinder = 2rh  2 
½m
22 7
  5  110 cm 2
7 2
1m
Reqd - area = (550 + 110 – 77) cm2 = 583 cm2
15.
½m
Fiqure
½m
(i)
30
 tan 45o  1  y  30
y
1m
(ii)

x
1
y
30
 tan 30 o 
 x 

 10 3

y
3
3
3

1m

½m

16.
Height of building is 10 3 m
Area of Sq. ABCD = 142 or 196 cm2
½m
Area of Small Sq. = 42 or 16 cm2
½m
Area of 4 semi circles  4. 1 3.14 (2) 2  cm 2
 2

 25.12 cm 2
 Reqd. area  (196 – 16 – 25.12) cm 2 


 154.88 cm 2
17.





1m
1m
The given quadratic eqn. can be written as
(k + 1)x2 – 2(k – 1)x + 1 = 0
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For qual roots 4(k – 1) 2 – 4(k  1)  0
or k 2 – 3k  0
 k  0, 3
 Non-zero value of k = 3 : Roots are
18.



1m
1 1
,
2 2
½+½ m
Number of redface cards removed = 6
1m
 Remaining cards = 46
19.
20
10
or
46
23
(i)
P (a redcard) =
(ii)
P (a facecard) =
(iii)
P (a card of clubs) =
1m
6
3
or
46
23
1m
13
46
1m
Getting x = 1, y = – 4 P(1, – 4)
z = – 1, t = 2
1m
R(– 1, 2)
½m
Area  PQR

1m
1
1 2 – 2  – 1 2  4  3 – 4 – 2   1  24
2
2
 12 sq.u.
20.
1½ m
Let a be the first term and d the common difference of the A.P.
S30 = 15 [2a + 29d] = 30a + 435 d
1m
S20 = 10 [2a + 19d] = 20a + 190 d
½m
S10 = 5 [2a + 9d] = 10a + 45 d
3 (S20 – S10) = 3 (10a + 145d) = 30a + 435d = S30
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SECTION - D
21.
22.
Correctly stated
Given, to Prove, Construction and correct figure
2m
correct Proof
2m
PR  PQ 
o
180 – 30

 PRQ   PQR 
2
 75o
SR | | QP and QR is a transversal   SRQ  75o
  ORQ   RQO  90o – 75o  15o
1m



1m
  QOR  180 – 2  15  150o   QSR  75o
o
1m
 RQS  180 o –  SRQ   SQR   30o
23.
1m
figure
1m
(i)
x
1
 tan 30o 
 y 3x
y
3
1m
(ii)
x 5
x 5
 tan 60 o  3 or
 3
y
3x
Writing the trigonometric equations


24.
3x  x  5
or x  2.5
Height of Tower  2.5 m





1½ m
½m
Money required for Ramkate for admission of daughter = Rs. 2500
A.P. formed by saving
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(i)
= 100, 120, 140, --- upto 12 terms
12
2 100  11 20  6 420
2
 Rs. 2520
Sum of AP (i) 

25.




1½ m
She can get her doughter admitied
½m
Value : Small saving can fulfill your big desires or any else
1m
Let the fraction be
x –3
x
½m
By the given condition, new fraction



x – 3 2 x –1

x2
x2
x – 3 x – 1 29


x
x  2 20
20 x – 3 x  2  x x – 1  29 x 2  2x


2
2


2
20 x – x – 6  x – x  29x  58x







½m
1m
or 11x2 – 98x – 120 = 0
or 11x2 – 110x – 12x – 120 = 0
(11x + 12) (x – 10) = 0
 The Fraction is
26.

1m
x = 10
7
10
1m
Let x m be the internal radius of the pipe
Radius of base of tank = 40 cm =
1m
2
m
5
Level of water raised in the tank = 3.15 or
315
100
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2.52 km/hour  1.26 km in half hour = 1260 m

1m
Getting the equation
π x 2 . 1260  π .
2 2 315
. 
5 5 100
1m
4
315
1
1
.


25 100
1260
2500
1
 x
m  2 cm
50
 x2 






1½
Internal diameter of pipe = 4 cm
27.
½m
Area ABC

1
– 4 – 4  5 – 3 – 5 – 8  0 8  4 
2

1
2
– 4  39 
35
2
1½ m
Area of ACD

1
– 4 – 5 – 6  0 6 – 8  5 8  5
2

109
2
 Area of Qurd. ABCD =
28.
1½ m
35 109

 72 sq.u.
2
2
1 m.
Volume of earth taken out after digging the well
 22

   3  3  21 cu.m  59 4 cu.m
 7

Let h be the height of the platform
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
27 × 11 × h = 594

h 
1m
594
27  h
1m
 Height of platform = 2 m
29.
i)
Number of numbers dividible by 3 or 5 in numbers 1 to 25
(3, 6, 9, 12, 15, 18, 21, 24, 5, 10, 20, 25) : their number is 12
P (divisible by 3 or 5) =
12
25
1m
1m
No. of favourable outcomes = 5
ii)
P (a Perfect square number) =
30.
Correct Construction
31.
3
4
29


x 1 x –1
4x – 1
5 1

25 5
(1, 4, 9, 16, 25)
1m
4m
[3 (x – 1) + 4 (x + 1)] [4x – 1] = 29 (x2 – 1)
1m
(7x + 1) (4x – 1) = 29 x2 – 29
1m
28x2 – 3x – 1 = 29 x2 – 29
or
x2 + 3x – 28 = 0
1m
(x + 7) (x – 4) = 0
 x = – 7, 4
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1m